$\mathrm { A } + 5.0 - \mu \mathrm { C }$ point charge is placed at the $0 \mathrm {~cm}$ mark of a meter stick and a $- 4.0 - \mu \mathrm { C }$ charge is placed at the $50 \mathrm {~cm}$ mark. What is the net electric field at the $30 \mathrm {~cm}$ mark? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \right.$. $m ^ { 2 } / C ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The net electric field at the 30 cm mark is the vector sum of the electric fields due to the two charges. The electric field due to the 5.0 μC charge is directed to the right and has a magnitude of:$$E_1 = k\frac{5.0	imes10^{-6}	ext{ C}}{(0.30	ext{ m})^2} = 4.49	imes10^5	ext{ N/C}$$The electric field due to the -4.0 μC charge is directed to the left and has a magnitude of:$$E_2 = k\frac{4.0	imes10^{-6}	ext{ C}}{(0.20	ext{ m})^2} = 9.0	imes10^5	ext{ N/C}$$The net electric field is:$$E = E_1 - E_2 = 4.49	imes10^5	ext{ N/C} - 9.0	imes10^5	ext{ N/C} = -4.51	imes10^5	ext{ N/C}$$The negative sign indicates that the net electric field is directed to the left.

A+5.0−μC\mathrm { A } + 5.0 - \mu \mathrm { C }A+5.0−μC

$\mathrm { A } + 5.0 - \mu \mathrm { C }$